違約機率校準檢定 - global test 由兩部分組成：第一部分為 level，探討真實的平均違約機率是否被高估；第二部分 shape，探討高低違約機率的表現情形。但 global test 與相關違約事件下的 level test 檢定尺度皆遠高於顯著水準 $\alpha$。本文先是針對相關違約事件，利用截斷分配使 level test 犯型一誤差機率更接近顯著水準，並提出虛無假設及對立假設為 $H_0: \theta \in \cup_{i=1}^2 \Theta_{i0}$ vs. $H_1: \theta \in \cap_{i=1}^2 \Theta_{i1}$ 的形式，引用交聯集檢定。更進一步透過 Liu \& Berger (1995, \textit{The Annals of Statistics}, 23, 1, 55-72) 建構齊一較強檢力檢定，改善檢定力。模擬結果顯示交聯集檢定與齊一較強檢力檢定的檢定尺度皆為 $\alpha$，且齊一較強檢力檢定的檢定力皆高於交聯集檢定。 / The calibration test of the PDs (probabilities of default) --- global test is twofold, the first part is the level test, which is about the mean of calibrated PDs. Second, the shape test is about whether a calibrated PD model differentiates correctly between low and high default probability events. In simulation results, we found that the type I error of global test is much greater than significant level $\alpha$, so is level test in correlation default events. In this study, firstly, we use the truncated level test to control previous error and suggest the hypothesis $H_0: \theta \in \cup_{i=1}^2 \Theta_{i0}$ vs. $H_1: \theta \in \cap_{i=1}^2 \Theta_{i1}$. Secondly, we introduce the intersection union test (IUT). Moreover, we construct an uniformly more powerful test (UMP test) by Liu \& Berger (1995, \textit{The Annals of Statistics}, 23, 1, 55-72). Simulation results show that the IUT and UMP test are size $\alpha$ tests, and the power of UMP test is greater than IUT.

波松分配（Poisson Distribution）常用在單位時間或是區間內，計算對有興趣之某隨機事件次數（或是已知事件之頻率），例如：速食餐廳的單位時間來客數，又或是每段期間內，某天然災害的發生次數，可以表示為某一特定事件X服從波松分配，若lambda為單位事件發生次數或是平均次數，我們稱lambda為此波松分配之母數，記作Poisson(lambda)，其中lambda屬於實數。 今天我們若想要探討由兩個服從不同波松分配抽取的隨機變數，如下列所述：令X={(X1,X2)}為一集合，其中Xi為X(i,1),X(i,2),...,X(i,ni)~Poisson(lambda(i)),i=1,2。欲探討兩波松分配之均數是否相同或相差小於某個常數d時，考慮以下檢定:H0:lambda2-lambda1<=d與H0:lambda2-lambda1>d，對於此問題可以使用的檢定方法有Przyborwski和Wilenski(1940)提出的條件檢定（Conditional test,C-test）或K.Krishnamoorthy與Jessica Thomson(2002)提出的精確性檢定（Exact test,E-test），其中的精確性檢定為一個非條件檢定（Unconditional Test）；K.Krishnamoorthy與Jessica Thomson比較條件檢定與精確性檢定的p-value皆小於顯著水準(apha)，而精確性檢定的檢定力不亞於條件檢定，因此精確性檢定比條件檢定更適合上面所述之假設問題。 Roger L.Berger（1996）提出一個以信賴區間的p-value所建立的較強力檢定，而目前只用於檢定兩二項分配（Binomial Distribution）的機率參數p是否相同為例，然而Berger在文中提到，較強力檢定比非條件檢定有更好的檢定力，而且要求的計算時間較少，可以提升檢定的效率。 本篇論文我們希望在固定apha與d時檢定的問題，建立一個兩波松分配均數顯著水準為apha的較強力檢定。 利用Roger L.Berger與Dennis D.Boos(1994)提出以信賴區間的p-value方法，建立波松分配兩兩母均數差的較強力檢定；研究發現此較強力檢定與精確性檢定的p-value皆小於apha，然而我們的檢定的檢定力皆不亞於精確性檢定所計算得出的檢定力，然而其apha及虛無假設皆需要善加考慮以本篇研究來看，當檢定為單尾檢定時，若apha<0.01，我們的較強力檢定沒有辦法找到比精確性檢定更好地拒絕域，換言之，此時較強力檢定與精確性檢定的檢定力將會相等。 / Poisson Distribution is used to calculate the probability of a certain phenomenon which attracted by researcher. If we want to test two random variable in an experiment .Therefore ,let X={(X1,X2)} be independent samples ,respectively ,from Poisson distribution ,also X(i,1),X(i,2),...,X(i,ni)~Poisson(lambda(i)),i=1,2. The problem of interest here is to test: H0:lambda2-lambda1<=d and H0:lambda2-lambda1>d, where 0<apha<1/2 ,and let Y1 equals sum of X1 and Y2 equals sum of X2, where apha ,lambda,d be fixed. In this problem of hypothesis testing about two Poisson means is addressed by the conditional test.However ,the exact method of testing based on the test statistic considered in K.Krishnamoorthy,Jessica Thomson(2002) also commonly used. Roger L.Berger ,Dennis D.Boos(1994) give a new way to calculate p-value,which replace the old method ,called it a valid p-value .In 1996, Roger L.Berger used the new way to propose a new test for two parameter of binomial distribution which is more powerful than exact test. In the other hand, Roger L.Berger also explain the unconditional test is more suitable than the conditional test. In this paper,we propose a new method for two parameter of Poisson distribution which revise from Roger L.Berger’s method. The result we obtain that our new test is really get a much bigger rejection region.We found when the fixed increasing ,the set of more powerful test increasing, and when the fixed power increasing ,the required sample size decreasing.